Question

已知函数f(x)=cos(x-2π/3)-cosx,求函数的最小正周期及单调递增区间

Answer 1

cos(x-2π/3)-cosx

=cosxcos(2π/3)+sinxsin(2π/3)-cosx

=-(3/2)cosx+(√3/2)sinx

=√3[sinx*cos(π/3)-cosx*sin(π/3)]

=√3sin(x-π/3)

（1）T =2π

（2） 2kπ-π/2 ≤x-π/3≤2kπ+π/2

2kπ-π/6 ≤x≤2kπ+5π/6

所以 增区间【2kπ-π/6 ,2kπ+5π/6】,k∈Z